Question

What is the relative decrease in focal length of a lens for an increase in optical power by \(0.1\ \text{D}\) from \(2.5\ \text{D}\)?

• A. \(0.04\)
• B. \(0.40\)
• C. \(0.1\)
• D. \(0.01\)

Hint:

Since \(P=\frac{1}{f}\), increase in power causes decrease in focal length. Relative decrease \(=\frac{P_2-P_1}{P_2}\).

Answer 1

The Correct Option is A

Concept:
Power of a lens is reciprocal of focal length in metre. \[ P=\frac{1}{f} \] Therefore, \[ f=\frac{1}{P} \] If power increases, focal length decreases.

Step 1: Write the initial and final power.
Initial power: \[ P_1=2.5\ \text{D} \] Increase in power: \[ \Delta P=0.1\ \text{D} \] Final power: \[ P_2=2.5+0.1=2.6\ \text{D} \]

Step 2: Write initial and final focal lengths.
Initial focal length: \[ f_1=\frac{1}{P_1} \] Final focal length: \[ f_2=\frac{1}{P_2} \]

Step 3: Find relative decrease in focal length.
Relative decrease is: \[ \frac{f_1-f_2}{f_1} \] Substitute \(f_1=\frac{1}{P_1}\) and \(f_2=\frac{1}{P_2}\): \[ \frac{\frac{1}{P_1}-\frac{1}{P_2}}{\frac{1}{P_1}} \] \[ =1-\frac{P_1}{P_2} \] \[ =1-\frac{2.5}{2.6} \] \[ =\frac{2.6-2.5}{2.6} \] \[ =\frac{0.1}{2.6} \] \[ =0.03846 \]

Step 4: Choose the nearest option.
\[ 0.03846 \approx 0.04 \] Hence: \[ \boxed{0.04} \]